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-BARSIC- [3]
3 years ago
13

A nonzero polynomial with rational coefficients has all of the numbers [1 sqrt{2}, ; 2 sqrt{3}, ;3 sqrt{4},; dots, ;1000 sqrt{10

01}]as roots. What is the smallest possible degree of such a polynomial?
Mathematics
1 answer:
Colt1911 [192]3 years ago
8 0

Answer:

Its degree can be at least 1970

Step-by-step explanation:

for each root of the form √q, where q is not a square, we have a root -√q. Therefore, we need to find, among the numbers below to 1000, how many sqaures there are.

Since √1000 = 31.6, we have a total of 30 squares:

2², 3², 4², ...., 30², 31²

Each square gives one root and the non squares (there are 1000-30 = 970 of them) gives 2 roots (one for them and one for the opposite). Hence the smallest degree a rational polynomial can have is

970*2 + 30 = 1970

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JulsSmile [24]
<h2>Part 1)</h2>

Let's analyze each part of the function:

FROM x = 0 TO x = 5:

We can find the equation of this line by using The Slope-Intercept Form of the Equation of a Line, that states:

The \ graph \ of \ the \ equation: \\ \\ y=mx+b \\ \\ is \ a \ line \ whose \ slope \ is \ m \ and \ whose \ y-intercept \ is \ (0, b)

So the y-intercept here is (0,4) and m=\frac{4}{5}, therefore:

\boxed{y=\frac{4}{5}x+4}

FROM x = 0 TO x = 5:

From the previous line, we know that at x=5 the output is:

y=\frac{4}{5}x+4 \\ \\ y=\frac{4}{5}(5)+4 \\ \\ y=4+4=8

So the point P_{1}(5,8) lies on both lines.

For this new line, the slope is m=-\frac{3}{5}

So, with the Point-Slope Form of the Equation of a Line we can find the equation of this other line:

The \ equation \ of \ the \ line \ with \ slope \ m \\ passing \ through \ the \ point \ (x_{1},y_{1}) \ is:\\ \\ y-y_{1}=m(x-x_{1})

So:

y-8=-\frac{3}{5}(x-5) \\ \\ y=8-\frac{3}{5}x+3 \\ \\ \boxed{y=-\frac{3}{5}x+11}

The graph is shown below.

<h2>Part 2)</h2>

The graph of the linear function f(x)=ax+b is a line with slope m=a and y-intercept at (0,b). From the items, we can assure that the following equations are linear functions:

\bullet 3y=2x-1.5 \ because \ y=\frac{2}{3}x-\frac{1}{2} \\ \\ \\ \bullet y=1 \ because \ y=m(0)+1 \\ \\ \\ \bullet 5(x+y)=-25 \ because \ 5x+5y=-25 \ \therefore 5y=-5x-25 \therefore y=-x-5

In conclusion, the other functions are nonlinear and they are:

\bullet \ y=x^2+3 \\ \\ \bullet \ y=x^3 \\ \\ \bullet \ y=12-x^2

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